Bootcamp

Christoph Thiele

Summer 2012

Contents

1 Metric spaces 11.1 Examples of metric spaces . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Baire argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.7 Total boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.8 Cauchy sequences, convergent sequences . . . . . . . . . . . . . . . . 151.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.10 Mappings between metric spaces . . . . . . . . . . . . . . . . . . . . . 171.11 Contraction mapping principle . . . . . . . . . . . . . . . . . . . . . . 221.12 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.13 The space C(M,M ′) . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.14 Arzela Ascoli . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.15 The Riemann integral for continuous functions . . . . . . . . . . . . . 30

1 Metric spaces

1.1 Examples of metric spaces

Definition 1 A metric space is a pair (M,d) consisting of a set M with a functiond : M ×M → R+

0 with the following properties that hold for all x, y, z ∈M

1. Zero distance for identical element: d(x, x) = 0.

2. Non-zero distance for non-identical elements: If x 6= y, then d(x, y) > 0

3. Symmetry: d(x, y) = d(y, x)

4. Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z)

1

Sometimes we refer to M as thye metric space, when the distance function d isassumed clear from the context.

Note that any subset M ′ of a metric space M is again a metric space, if thedistance function is restricted to M ′ ×M ′ from M ×M .

We list some examples of metric spaces, by the above remarks any subset of thesespaces is again a metric space.

1. D with distance d(x, y) = |x− y| =√

(x− y)2

2. R with distance d(x, y) = |x− y| =√

(x− y)2

3. Rn, i.e. the set of tuples x = (x1, . . . , xn) with xi ∈ R with distance d(x, y) =√∑ni=1(xi − yi)2

4. Rn with distance d(x, y) = sup1≤i≤n |xi − yi|

5. l2(N), the set of sequecnes x : N→ R with distance d(x, y) =√∑∞

i=1(xi − yi)2

6. l∞(N), the set of bounded sequences, with distance d(x, y) = supi∈N |xi − yi|

7. l1(N), the set of absolutely summable sequences, with d(x, y) =∑∞i=1 |xi − yi|

8. The set BV (D) of functions of bounded variation on D, i.e. the set of pairs(f, g) with f, g right continuous (say) monotone increasing bounded functionsmodulo the equivalence relation (f, g) ∼ (f ′, g′) if f + g′ = f ′ + g with distance

d((f, g), (h, k)) = inf[| supx

(f ′+k′)(x))−(f ′+k′)(0)|+| supx

(g′+h′)(x)−(g′+h′)(0)|]

where the infimum is taken over all representatives of ther respective the equiv-alence classes.

9. The space C([0, 1]) of continuous functions f : [0, 1]→ R with distance

supx∈[0,1]

|f(x)− g(x)|

10. The space C(R+0 ) of continuous functions f : R+

0 → R with distance

supx∈R+

0

|f(x)− g(x)|

11. The space C0(R+0 ) ⊂ C(R+

0 ) of functions satisfying

lim supx→∞

|f(x)| = 0

with the same distance.

2

We leave the verification of the metric space properties in most of these cases as anexercise, just elaborating on some specific examples. We show the triangle inequalityin the case of l2(N). Let xi and yi two sequences with

∑i x

2i < ∞ and

∑i y

2i < ∞.

We frist claim the inequality

∑i

|xiyi| ≤√

(∑i

x2i )(∑i

y2i )

Note that this inequality is homogeneous, if we multiply xi or yi by a positive realnumber, then the inequality remains true. Hence it suffices to prove this inequalityunder the assumption ∑

i

x2i =∑i

y2i = 1

But then the inequality becomes equivalent to

2∑i

|xiyi| ≤ 2 = (∑i

x2i ) + (∑i

y2i ) =∑i

(x2i + y2i )

But then the inequality follwos by the comparison principle since for each i

2|xiyi| ≤ x2i + y2i

which follows from0 ≤ (|xi| − |yi|)2

by adding 2|xiyi| on both sides.We thus conclude the Cauchy Schwarz inequality

∑i

xiyi ≤√

(∑i

x2i )(∑i

y2i )

But then we have ∑i

(xi + yi)2 =

∑i

x2i +∑i

y2i + 2∑i

xiyi

≤∑i

x2i +∑i

y2i + 2√

(∑i

x2i )(∑i

y2i ) = (√∑

i

x2i +√∑

i

y2i )2

Upon setting xi = ai− bi and yi = bi− ci this becomes the triangle inequality for thesequences ai, bi, ci

We also show the triangle inequality for the space BV (D). For simplicity ofnotation we restrict attention to the space of functions therein which vanish at 0,which can be expressed using pairs of functions vanishing at 0. Then the distance isexpressed as

d((f, g), (h, k)) = inf[supx

(f ′ + k′)(x) + supx

(g′ + h′)(x)]

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where again the infimum is taken over all choices of representatives of the given class.(exercise: do the reduction of the general case to this case.)

Let (f, g), (h, i), (j, k) be three pairs of functions. Assume that the first two arechosen appropriate representatives so that

d((f, g), (h, i)) + ε ≥ supx

(f + i)(x) + supx

(g + h)(x)

Assume that (j, k) is chosen so that for some equivalent pair h′, i′ to h, i

d((j, k), (h, i)) + ε ≥ supx

(j + i′)(x) + supx

(k + h′)(x)

Then we have h+ i′ = h′ + i and (f, g) is equivalent to (f + i+ h′, g + h+ i′). Hence

d((f, g), (j, k)) ≤ supx

(f + i+ h′ + k)(x) + supx

(g + h+ i′ + j)(x)

≤ supx

(f + i)(x) + supx

(g + h)(x) + supx

(h′ + k)(x) + supx

(i′ + j)(x)

d((f, g), (h, i)) + ε+ d((j, k), (h, i)) + ε

Since ε was arbitrary, this proves the triangle inequality.

Approximation in metric spaces

We begin by noting some similarity of the metric space axioms to the axioms of anequivalence relation. Asd a matter of fact, we have the following lemma:

Lemma 1 Assume a M is a set and d a function M×M → R+0 satisfying properties

1, 3, 4 of a metric space. Then the relation defined by x ∼ y if d(x, y) = 0 is anequivalence relation on M .

Proof: For all x ∈ M we have d(x, x) = 0 by the first property and thus x ∼ xwhich imlies reflexivity. Symmetry follows from symmetry of d. Transitivity: If x ∼ yand y ∼ z then d(x, z) ≤ d(x, y) + d(y, z) = 0 + 0 = 0. hence x ∼ z. 2

One can use thsi equivalence relation to pass to the space of equivalence classes,which then becomes a metric space.

Lemma 2 Assume a M is a set and d a function M×M → R+0 satisfying properties

1, 3, 4 of a metric space. Let M be the set of equivalence classes with respect to theequivalence relation x ∼ y if d(x, y) = 0. Then the function d from M × M → R+

0

defined byd([x], [y]) = d(x, y)

is well defined, and makes (M, d) a metric space.

4

Proof: We need to prove that the distance function depends only on the equiva-lence class [x] and [y] of x, y. Let x ∼ x′ and y ∼ y′. Then by the tirangle inequality

d(x, y) ≤ d(x, x′) + d(x′, y′) + d(y′, y) = d(x′, y′)

where we have used the similarity relation in the last identity. Similarly we havethe reverse inequality. Thus d(x, y) = d(x′, y′). Finally we show that M is a metricspace with metric d. Properties 1, 3, 4 follow immediately form the correspondingproperties of M . To verify Property 2, observe that if x and y are not equivalent,then d(x, y) 6= 0 and hence d(x, y) 6= 0. 2

One may make an attempt to take the the above insight further saying that therelation x ∼ y if d(x, y) ≤ ε is almost an equivalence relation. It is reflexiv andsymmetric, but fails to be transitive by a small error: if d(x, y) and d(y, z) ≤ ε, thenby the triangle inequality d(x, z) ≤ 2ε, which is still small though not sufficientlysmall to guarantee equivalence. This almost equivalence relation is behind the ideaof approximation in metric spaces. We define almost equivalence classes, called balls:

Definition 2 (Ball) Let (M,d) be a metric space. For x ∈M and r ∈ D we definethe ball with center x and radius r to be

Br(x) = {y ∈M : d(x, y) < r}

The empty set is a ball with radius r = 0.

Note that the balls encode the full information of the metric space: knowing whichelement is in which ball is enough to determine the function d.

A dense set may be considered a sufficient set to pick a representtaive out of eachalmost equivalence class:

Definition 3 (Dense) Let (M,d) be a metric space. A subset M ′ of M is calleddense in M , if every ball Br(x) in M contains at least one point of M ′.

1.2 Cones

Given a metric space (M,d), we call M ×D the upper half space of the metric space.For each point x ∈M we can define a cone

C(x) = {(y, r) ∈M ×D : d(y, x) ≤ r}

Thus the horizontal section of the cone at height r is the intersection of all balls ofradius s about x with s > r.

Definition 4 A pre-cone S is a subset of the upper half space satisfying

1. For each r > 0 there is 0 < s ≤ r and a y ∈M such that (y, s) ∈ S

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2. For each pair (x, r), (y, s) ∈ S we have the inequality

d(x, y) ≤ r + s

By the triangle inequality, the cone C(x) is a pre-cone. Namely, for all (y, s), (z, t) ∈S we have

d(y, z) ≤ d(y, x) + d(x, z) ≤ s+ t

Cones are pre-cones that are maximal in the sense:

Definition 5 A pre-cone C is called a cone if for all (y, s) /∈ C there is a (z, t) ∈ Cwith

d(y, z) > s+ t

Lemma 3 For each pre-cone, there is a unique cone containing the pre-cone.

Proof: Existence: define C to be the set of all (x, r) ∈ M ×D such that for all(y, s) ∈ S we have

d(x, y) ≤ r + s

By definition of the pre cone we have S ⊂ C. Then we have for all (x, r), (x′, r′) in Cand all (y, s) ∈ S

d(x, x′) ≤ d(x, y) + d(y, x′) ≤ r + s+ s+ r′

Now there exists (y, s) ∈ S with arbitrarily small s, hence

d(x, x′) ≤ r + r′

Hence C is a pre cone. If (y, r) /∈ C then by definition there exists (y, s) ∈ S suchthat

d(x, y) > r + s

Since S ⊂ C, this proves that C is a cone.Uniqueness: Let C ′ be another cone containing S. Since S ⊂ C ′ and for each

(y, s) ∈ C ′ we haved(x, y) ≤ r + s

for each (x, r) ∈ S, the above argument shows that (y, s) ∈ C. Note that if (z, t) ∈ C,then since C ′ ⊂ C we have

d(y, z) ≤ s+ t

for each (y, s) ∈ C ′, hence (z, t) ∈ C ′. Hence C = C ′ 2.

Lemma 4 A nested family of balls is a family of balls Br(xr) parameterized by r in asubset A ⊂ D such that Br(xr) ⊂ Bs(xs) whenever r ⊂ s. Given such a family whereA contains arbitrarily small positive elements, then the set

C = {(z, 2t) : z ∈ Bt(xt)}

is a pre cone.

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Proof: Let (y, s), (z, t) ∈ C and assume without loss of generality that s < t.Then we have:

d(y, z) ≤ d(y, xs) + d(xs, z) ≤ 2t+ s ≤ 2t+ 2s

2

Definition 6 A tip of a pre-cone C is an element of the form (x, 0) ∈ C

Lemma 5 There is at most one tip of any pre-cone.

Proof: Let (x, 0) and (y, 0) be two tips, then by the cone property we have

d(x, y) = 0 + 0 = 0

which proves x = y. 2

1.3 Completeness

A pre-cone may or may not contain a tip.

Definition 7 A metric space is called complete, if every pre-cone has a tip.

A cone is the natural pendant for metric spaces of a Dedekind cut for D. Thissuggest we may seek an analoguous process in general metric spaces for the extensionsfrom the metric space D to the metric space R+

0 .

Lemma 6 Let (M,d) be a metric space. The set M of cones of this metric spacewith distance d(C,C ′) defined as the infimum of all numbers c such that

d(x, y) ≤ r + c+ s

for all (x, r) ∈ C and (y, s) ∈ C ′ is a metric space.

Proof: We first show that the infimum is not∞ and hecne in R+0 . Fix (x′, r′) ∈ C

and (y′, s′) ∈ C ′. Then for all (x, r) ∈ C and (y, s) ∈ C ′ we have

d(x, y) ≤ d(x, x′) + d(x′, y′) + d(y′, y) ≤ r + r′ + d(x′, y′) + s′ + s

This impliesd(C,C ′) ≤ r′ + d(x′, y′) + s′

which shows that this distance is finite.If C = C ′, then we have

d(x, y) ≤ r + 0 + s

for all (x, r), (y, s) ∈ C. hence equal cones have zero distance. Assume d(x, y) = 0.Fix (x, r) ∈ C, then we have for all ε > 0 and (y, s) ∈ C ′

d(x, y) ≤ r + ε+ s

7

Since ε was arbitrary, we have d(x, y) ≤ r + s. Since (y, s) ∈ C ′ was arbitrary, wehave (x, r) ∈ C ′ and hance C ⊂ C ′. By symmetry we also have C ′ ⊂ C, and henceC = C ′. Symmetry of the distance d is clear by definiton.

Now let C,C ′, C ′′ be cones. Then we have for every (x, r) ∈ C, (y, s) ∈ C ′ and(z, t) ∈ C ′′

d(x, z) ≤ d(x, y) + d(y, z) ≤ r + d(C,C ′) + s+ s+ d(C ′, C ′′) + t

Since C ′ contains poitns with arbitrarily small s, we conclude

d(x, z) ≤ d(x, y) + d(y, z) ≤ r + d(C,C ′) + d(C ′, C ′′) + t

By definition of the infimum, this implies

d(C,C ′′) ≤ d(C,C ′) + d(C ′, C ′′)

2

Lemma 7 Let (M,d) be a metric space and (M, d) be the space of cones of M . Thenthere is an isometric embedding from M to M sending x to the cone C(x). The rangeof this map is dense in M

Proof: Let x, y ∈M . Then for (x′, r) ∈ C(x) and (y′, s) ∈ C(y) we have

d(x′, y′) ≤ d(x′, x) + d(x, y) + d(y, y′) ≤ r + d(x, y) + s

hecne d(C(x), C(y)) ≤ d(x, y). On the other hand we have (x, 0) ∈ C(x) and (y, 0) ∈C(y) and

d(x, y) = 0 + d(x, y) + 0

shows that d(C(x), C(y)) ≥ d(x, y). hence the above map is an isometry. It istherefore also injective, and thus an embedding.

To see density, let C be a cone in M and let r > 0. By definition of the cone thereis x ∈M with (x, s) ∈ C and 0 < s < r. Then for any (z, t) ∈ C we have

d(z, x) ≤ t+ s = t+ s+ 0

and henced(C,C(x)) ≤ s

This proves density. 2.

Lemma 8 With the above setup, the space (M, d) is complete.

Let ι be the embedding from M to M . Let C be a cone in M . Let S = C∩ ι(M)×D.By density of M for each r > 0 there exists x ∈M with (C(x), s) ∈ C with 0 < s < r.Hence S is a pre-cone. Let S be the subset of M × D of all points (x, r) with(ι(x), r) ∈ S. Since ι is an isometry, S is a pre-cone. Let C be the associated cone.We claim C = C(C) 2

8

Exercise 1 Let (M,d) be a complete metric space. Let M ′ be any subset of M . Thenthe embedding map i : M ′ → M extends to the completion of M ′. Any cone in Mwhich has arbitrarily low elements in M ′ has a tip in the image of the completion.

Exercise 2 Consider a complete metric space. Let Bt(xt) be a set of nested ballsparameterized by t ∈ A where A ⊂ D. Then there exists an element x ∈ M which iscontained in the balls B2t(xt).

Exercise 3 Let (M,d) be a complete metric space and M ′ be a subspace. If and onlyif M/M ′ is the union of balls, then M ′ is complete.

1.4 Baire argument

Lemma 9 Let (M,d) be a nonempty complete metric space. Let Mn be a sequenceof subspacess such that each Mn is a union of balls and dense in M . Then tyheintersection of the spaces Mn is not empty.

Proof: Let Br0(x0) be a ball contained in M0. Assume we have already chosenBrn(xn). Since An+1 is dense, it has nonempty intersection with Brn(xn). Let Bn bea ball contained in An+1 that has nonempty intersection with Brn(xn). Then thereexists a ball Brn+1(xn+1) contained in Brn(xn)∩Bn. Choose this ball, we may assumethat its double ball is contained in Brn(xn)∩Bn. This gives a nested sequence of balls.By completeness, the associated pre-cone has a tip, which is contained in B2rn+1(xn+1)which in turn is contained in Brn(xn) which in turn is contained in An. hecne theintersection of the An is not empty. 2

Definition 8 A set X ⊂ M is called nowehre dense if its complement is dense andthe union of balls.

As a corollary of the above lemma, a complete metric space is not the union ofnowhere dense sets.

Definition 9 A point x in a metric space M is called isolated, if there exists r > 0such that Br(x) = {x}.

Lemma 10 A complete metric space without isolated points is uncountable.

Proof. For each x ∈M . the set M \ {x} is dense. Namely, since x is not isolated,the ball Br(x) contains some point of M \ {x}. For balls centered at points otherthan x, these balls trivially contain soem point in the set. The set M \{x} is alos theunion of balls, since for every y 6= x the bal Bd(x,y)(y) is contained in M \{x}. Assumeto get a contradiction that M is countable. Then M is the union of countably manysets of the above type. But this contradicts the previous lemma. 2

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1.5 Separability

Definition 10 Let (M,d) be a metric space and let M ′ be a subset of M . A set Cof balls in M is called a cover of M ′, if

M ′ ⊂⋃B∈C

B

A subset C ′ ⊂ C is called a subcover, if

M ′ ⊂⋃B∈C′

B

A cover is called countable if it consists of countably many or balls. Our use ofthe phrase “countable set” includes the possibility that the set is finite.

A subset C ′ of a cover C is called subcover if it is itself a cover of the subspace inquestion.

Definition 11 (Separable) A metric space (M,d) is called separable, if every coverof M has a countable subcover.

One could interpret this definition as follows: no matter how one chooses almostequivalence classes, countably many are enough.

Lemma 11 Countable metric spaces are separable.

Proof: Assume we are given an arbitrary cover. For each element in the separablespace choose one ball of the cover that contains the element. This gives a countablesubcover. 2

Lemma 12 If a metric space has a countable dense set, then it is separable.

Proof: Assume M has a countable dense subset M ′. Consider the set C of allballs Br(x) with r ∈ D and x ∈M ′. This is a countable set. Now let C ′ be any coverof M . Let C ′′ be the set of all balls in C which are contained in some ball of C ′. Weclaim C ′′ is a cover of M . Namely, let x ∈ M . then there exists Br(y) ∈ C ′ whichcontains x. Then Bs(x) ∈ Br(y) for some s > 0. Then there exists z ∈M ′ containedin Bs/2(x) By the triangle inequality Bs/2(z) ⊂ Br(y). Hence Bs/2(z) ∈ C ′′ and C ′′

is a cover. For each ball Br(x) in C ′′ choose one ball in C ′ which contains the ballBr(x). the set of chosen balls is countable and covers M . 2

Lemma 13 If a metric space is separable, then it has a countable dense set.

Proof. for each r > 0 consider the set M of all balls of radius r. Clearly every pointx ∈M is contained in the ball Br(x), so these balls cover M . By assumption there isa countable subcover Cr. Let Xr be the set of centers of balls in Cr and let X be thecountable union of the Xr. Then X is countable. We claim it is dense. Let Bs(y) beany ball in M . Since Cs is a cover, there is a ball Bs(z) ∈ Cs which contains y. Butthen z ∈ Bs(y). This proves that X is dense. 2.

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Exercise 4 Prove any number of the following:

1. l2(N) is separable.

2. BV (D) is not separable.

Lemma 14 Any subset of a separable metric space is separable.

Note that given any countable dense subset of M , it is not clear that this set hasany point in common with M ′. hence construction of a countable dense subset of M ′

has to be more careful.Proof: Since M is separable, we find a countable dense subset X. Consider the

countable collection C of all balls with centers in X and radius in D. For eachball Br(x) in C such that intersects M ′, pick a point in Br(z) ∩M ′. This gives acountable collection Y of points in M ′. We claim that this collection is dense in M ′.Pick z ∈ M ′ and s > 0. Since Cs/2 covers M , there is a ball in Cs/2 which containsz. This ball contains a chosen point y ∈ Y as well, and we have by the triangleinequality y ∈ Bs(z). This proves density of Y . 2

Conversely we have

Lemma 15 Let M ′ be a metric subspace of a metric space M . Assuyme we have acover of M ′ by balls in M , then there exists a countable subcover.

Note that this lemma is not immediate from the definiiton of separability, sincethe balls in the given cover may not have centers in M ′ and thus their restrictions toM ′ may not be balls in M ′.

Proof: For each point y ∈M ′ there is a ball Br(x) in the cover containing y. Thenthere is a small ball Bs(y) contained in Br(x). The collection of all such small ballsfor y ∈M ′ covers M ′ and their restrictions to M ′ are a cover in the space M ′. Hencethere is a finite subcover. For each ball Bs(y) in this subcover pick one of the bigballs Br(x) as above. This gives a countable subcover of the original cover. 2

Definition 12 Let (M,d) and (M ′, d′) be two metric spaces. The Cartesian productof the two metric spaces is the set of all ordered pairs (x, x′) with x ∈M and x′ ∈M ′

and distanced((x, x′), (y, y′)) = max{d(x, y), d(x, y′)}

Exercise 5 Prove that this is indeed a metric space.

Exercise 6 The Cartesian product of two separable metric spaces is separable.

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1.6 Compactness

Definition 13 A metric space is called compact, if every cover of the space has afinite subcover.

As in the case of separability we have the following two observations:

Lemma 16 Finite sets are compact.

Lemma 17 If a metric subspace M ′ of a space M is compact and we are given anycollection of balls in M that cover M ′, then there is a finite subcover.

The analogy of the theories of separable and compact sets has an end when thecountability of the set of radii D comes into play. For example, compact spaces donot have finite dense sets (finite sets are never dense in any larger space). Also, asubset of a compact set need not be compact.

On the other hand one has some immediate consequences from applying the com-pactness property to all balls centered at a given point:

Definition 14 A metric space (M,d) is bounded if there exists a ball which containsthe whole space M .

This is clear if M is empty by using the empty ball.

Lemma 18 If M is a metric space and M ′ a subspace, and if M ′ is contained in aball with center in M , then M ′ is bounded.

Proof: We need to show there exists a ball with center in M ′ which contains M ′.We may assume M ′ is nonempty. Let Br(x) a ball with center in M which containsM ′. Let y be any point in M ′. Then B2r(y) contains M ′ by the triangle inequality.2

Lemma 19 A compact metric space (K, d) is bounded.

Proof: We may assume K is not empty. Pick a point x ∈ K and observe that

M =⋃r∈D

Br(x)

, since every point y in K has some finite distance d(x, y) to x and there exists r ∈ Dwith d(x, y) < r and hence y ∈ Bt(x). Hence the collection of balls about x coversthe compact set, and hence there is a finite collection of balls about x which coverthe compact set. Since the balls are nested, it suffices to take the largest ball to coverthe compact set. This proves boundedness. 2

There is a dual observation using complements of balls

Lemma 20 Every compact metric space (K, d) is complete.

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Proof: Pick a cone B in K. For every (x, r) /∈ B there exists ε > 0 and (y, s) ∈ Bsuch that

d(x, y) ≥ r + s+ ε

Pick the ball Bε(r) and let C be the collection of all these balls.We have two cases:

1. C covers K. Then there is some finite subcover. Each ball in the cover comeswith a point (y, s) ∈ B as above, pick s0 the minimal s. If s = s0, we are donesince then the cone has a tip. We claim s0 > 0 is impossible. For assume to geta contradiction that s0 > 0. Pick (z, t) ∈ B with t < s0. The point z is in thefinite subcover. Let B be the ball in the cover containing z , it comes with apoint (x, y) /∈ B and (y, s) ∈ B with

r + s+ ε ≤ d(x, y)

Thenr + t+ ε ≥ d(x, z) + d(y, z) ≥ d(x, y) ≥ r + s+ ε

This is a contradiction since t < s0 ≤ s.

2. Assume C does not cover K, and let z be a point not covered. Then there doesnot exist (y, s) ∈ B with

d(z, y) ≥ r

Hence the cone is of the form B(z)

Lemma 21 Let (M,d) be a compact metric space and M ′ be a subspace. If and onlyif M/M ′ is the union of balls, then M ′ is compact.

Proof: Let C be a cover of M ′. Since M \M ′ is the union of balls, the set C ′ ofthese balls forms a cover of M \M ′. Hence C ∪C ′ covers M . Find a finite subcover.Remove all balls in C ′ from this finite subcover, then it still covers M ′

The only if part follows form only if for completeness. 2

Lemma 22 Let K be a compact metric space and let Kn be a nested sequence ofcompact subsets of K, i.,e., Kn+1 ⊂ Kn. Then the intersection of the sets Kn is notempty.

Proof: the complement of Kn is a union of open balls by the previous lemma.Take the collection of all such balls for all n. If these balls cover all of K, there is afinite subcover. This finite subcover is disjoint from some Kn. Thsi is a contradiction.Hence these balsl do not cover K, hence the Kn have nonempty intersection.

The following is a version of the fact that finite subsets in a totally ordered spacehave a maximum and a minimum.

13

Lemma 23 Let (M,d) be a metric space and let K be a subspace. Let x ∈ M , thenthere exists y and z in K such that

d(x, y) = inft∈K

d(x, t)

d(x, y) = supt∈K

d(x, t)

Proof: There is a sequence tn of elements in K such that d(x, tn) is monotoneincreasing and

supnd(x, tn) = sup

t∈Kd(x, t) = S

Pick a subsequence that is Cauchy. Then this seqeunce has a limit t ∈ K. Then wehave

d(x, t) ≥ d(x, tn)− d(t, tn) ≥ S − ε− ε

for arbitrarily small ε. Henced(x, t) = S

The proof of existence of a minimum is similar. 2

Lemma 24 The Cartesian product of two compact metric spaces is compact.

Proof: It suffices to find finite subcovers of covers by balls in this metric. Suppose weare given such a cover. For each x ∈ M this covers in particular the set {x} ×M ′.Find a finite subcover. Pick the minimal radius r of this finite subcover. Then thisfinite subcover covers Br(x) ×M ′. Now these Br(x) cover all of M . Find a finitesubcover. This produces a finite colelction of ifnite sets covering all of M ×M ′ 2

1.7 Total boundedness

The following definition is a property between separability and compactness.

Definition 15 A subset M ′ of a metric space M is called totally bounded, if for everyr > 0 there exists a cover of M by balls of radius r.

Lemma 25 A totally bounded space is separable.

Proof: For each r pick a finite cover Cr of the space M . Let C be any cover of M .Let x be a point in M . and Br(y) a ball in C containing x. then there is a ball Bs(x)contained in Br(y). Pick a ball Bx in Cs/2 containing x, then this ball is contained inBr(y). The collection of all Bx is countable since contained in the countable union offinite sets Cr. Picking for each ball Bx a ball in C that contains Bx gives a countablesubcover of C. 2

Lemma 26 A compact metric space is totally bounded.

14

Proof: Apply the definiiton of compactness to the collection of balls of radius r. 2

Lemma 27 Let (K, d) be a totally bounded complete metric space. Then K is com-pact.

Proof:Denote by Cr a finite cover of M by balls of radius r. Pick a ball Br0(x0) in C1

that is not covered by finitely many balls in C. Such ball exists, since if each of thefinitely many balls in C1 where covered by finitely many balls in C, then there was afinite subcover of C for M .

Assume we have already selected the ball Brn(xn). Pick one ball Brn+1/2(xn+1) inCrn/2 that intersects Brn(xn) and is not covered by finitely many balls in C. Suchball exists since Cr/2 is a finite cover of M and since Brn(xn) is not covered by finitelymany balls in C.

The sequecne of balls B4rn(xn) is nested. Let x be the tip of the associated precone.It is contained in each ball B8rn(xn).

Let Br(y) be a ball in C that contains x. Then this ball contains a ball Bs(x).Then this ball contains the selected ball B8rn(xn) provided 16rn ≤ s, which can beachieved for large enough n. This however contradicts the fact that Brn(xn) is notcovered by finitely many balls of C.

2

Exercise 7 let M be a metric space and M ′ be a subspace. If for each r the spaceM ′ is covered by finitely many balls of radius r with center in M , then also for eachr it is covered by finitely many balls of radius r with center in M ′.

Exercise 8 Let M be a totally bounded metric space and assume there is a sequencef : N → M such that for any two balls Br(x) and Bs(y) with B2r(x) ∩ B2s(x) = ∅at most one of them contains infinitely many point of the sequence. Then there isexactly one cone B with the property that for every r there are at most finitely manypoints of the sequence not in {x : (x, r) ∈ C}

1.8 Cauchy sequences, convergent sequences

Minimal examples of pre cones are called Cauchy sequences.

Definition 16 Let (M,d) be a metric space. A sequence f : N → M is calledCauchy, if for every r > 0 there exists n ∈ N such that

supk≥0

d(f(n), f(n+ k)) ≤ r

Definition 17 Let (M,d) be a metric space. A sequence f : N → M is said toconverge to x if for every r > 0 there exists n ∈ N such that

d(f(n), x) ≤ r

15

Exercise 9 If a sequence converges to a point x, then it is Cauchy.

Lemma 28 Let f : N→M be a Cauchy sequence. Then there is exactly one cone Csuch that for every r the set of points in the seuquence that are not in {x : (x, r) ∈ C}is finite. If f converges, it converges to the tip of the cone.

Exercise 10 A metric space is complete if and only if every cauchy sequecnes con-verges.

Lemma 29 A metric space M is toatlly bounded, if and only if every sequence in thespace has a subsequence which is Cauchy.

Proof: First assume the space is totally bounded. Let f : N→M be a sequence .define f0 = f . Assume we have defined fn for n ≥ 0. Define a subsequecne fn+1 of fnso that fn+1(m) coincides with f n(m) for m ≤ n, and so that fn+1(m) ∈ B2−n(x) forsome ball of radius 2−n. This is possible, since we can cover the space M by finitelymany balls, so at least one of them needs to contain infinitely many values of thesequence fn. Then define g(n) = fn(n). This is a subsequecne of f , and it is Cauchy.

Now assume every sequecne has a subsequence that is Cauchy.Proof: Fix r and pick a point x0 ∈M . Assume we have already picked xn. If the

collection of balls Br(xk) with 0 ≤ k ≤ n covers M , we are done having constructed afinite cover of M . If not, pick xn+1 not covered by these balls. We claim this processcannot go on indefinitely. If it does, then we have a seuqence xn which has no Cauchysubsequecne since any two elements in the seuqence have distance at least r. this isa contradiction. 2

Exercise 11 A metric space is compact if and only if ever sequence has a convergentsubsequence.

Exercise 12 Let f : N → M be a sequence in a compact space. and let x ∈ M Ifevery subsequence of f has a further subsequence which converges to x, then f itselfconverges to x.

1.9 Examples

Note that Rn with Euclidean metric is complete.

Lemma 30 Bounded complete subsets of Rn with Euclidean metric are compact.

Proof: We prove that every sequence in K has a subsequence which is Cauchy.let (x1, . . . , xn) denote the sequence. Since the sequence is bounded, each entry isbounded. Choose a subsequence so that x1 is monmotone. Then choose a subsequenceso that x2 is monotone, and so on. Hence we may assume all sequences are monotone.But then all coordinate sequences are Cauchy, and hence the full sequence is Cauchy.Since K is complete, it has a limit.

2

16

Exercise 13 The space l2(N) is separable and complete.

Lemma 31 The ball B1(0) in l2(N) is not compact.

Proof: Let fn be the element in l2(N) which satisfies fn(m) = 0 if n 6= m andfn(m) = 1 if n = m. Then Any two elements have distance > 1, hence this sequencehas no subsequence that is Cauchy.

Exercise 14 Let a : N → R+0 be a sequence with lim supn→∞ an = 0. Let K be the

set of all sequences f in l2(N) (with values in R) such that

∞∑n=0

a(n)f(n)2 <∞ <∞

Prove that K is compact.

Exercise 15 Prove the converse of the above. If K ⊂ l2(N) is compact, then thereexists a sequence an as above such that

∑∞n=0 a(n)f(n)2 <∞ for all f ∈ K

1.10 Mappings between metric spaces

We study mappings f : M → M ′ between two metric spaces (M,d) and (M ′, d′)that in one way or the other respect the metric. However canonical the definitionof a metric space appeared, there are now a whole variety of classes of mappings toconsider. Essentially, one would like to have an upper bound for the distance of thefunction values in terms of the original distance.

A mapping f : M →M ′ is called:

1. An isometry, if d′(f(x), f(y)) = d(x, y) for all x, y

2. A contraction, if d′(f(x), f(y)) ≤ d(x, y) for all x, y

3. A Lipshitz map, if there exists L (Lipschitz constant) such that we have d′(f(x), f(y)) ≤Ld(x, y) for all x, y

4. A strict contraction if it is Lipshitz with constant L < 1

5. A Holder map with exponent 0 < α if there exists a constant C such thatd′(f(x), f(y)) ≤ Cd(x, y)α for all x, y The case α = 1 is that of a Lipschitz map.

6. A uniformly continuous map, if there exists a right continuous (convex?) mono-tone increasing function g : R+

0 → R+0 with g(0) = 0 such that g(d′(f(x), f(y))) ≤

d(x, y) for all x, y. The function g is called the modulus of continuity of thefunction. In case of a Holder function with exponent alpha this finction is ofthe form cx1/α for some small c.

17

7. A continuous map, if for every x ∈ M there exists a function gx as above suchthat for all y we have gx(d

′(f(x), f(y))) ≤ d(x, y) for all x, y. The function g iscalled the modulus of continuity of the function at x.

Exercise 16 Any isometry is injective.

Isometries are therefore often called isometric embeddings, they are bijective ontotheir range, and they describe a complete metric isomorphism between M and itsrange.

The example of isometries also suggests why one does not look at maps corre-sponding to the revers inequalities in the above list. these maps are informally ofan expanding nature, but they are injective in all these cases. Thus one may simplystudey the inverse map on the range and obtain a map of the above list. Thus onedoes not expect much new insight from studying reverse inequalities.

The class of contraction mappings plays a role in fixed point theorems, we willshow the classical fixed point theorem for strict contractions.

Contractions still do not have an interesting class of isomorphisms, if a bijectivemap is a contraction and its inverse is a contraction, then the map is already anisometry.

Note that for a metric space (M,d), the metric space (M,λd) for λ > 0 behavesvery similarly, except distances are measured in another unit. If one tries to look atequivalence relations of metric spaces under this scaling, then the notion of contractionmapping makes no sense anymore. Then the class of Lipschitz maps is the immediategeneralization of contraction mappings that is invariant under this scaling. NowLipschitz maps allow for inetresting non-trivial isomorphisms. A bijective map isbi-Lipschitz if both the map and its inverse are Lipschitz. There is a very rich classof bi-Lipshchitz maps, studied extensively. It is the most natural and most simplenotion of non-trivial isomorphisms of metric spaces if one wants to consider globalproperties of metric spaces and not just infinitesimally small features of the space.

Holder functions appear prominently in nature, for example the Brownian pathof a particle is almost surely Holder continuous with exponent α for any α < 1/2 butnot Holder continuous with exponent α = 1/2.

Lemma 32 Functions f : D → D which satsify Holder condition with α > 1 areconstant.

Proof: Let C be the the best constant in the Holder estimate for any three pointsx, y ∈ [0, 1]. We need to show that C = 0. Assume to get a contradiction C > 0.Consider a pair of points such that

C(1−−ε)|x− y|α < |f(x)− f(y)|

Then clearly x 6= y. Let z be the midpoint of x and y, then

|f(x)− f(y)| ≤ |f(x)− f(z)|+ |(f(z)− f(y)| ≤ C|x− z|α + C|z − y|α

18

C2(1

2)α|x− y|α

This however is a contradiction if

1− ε > 21−α

Since we could choose ε arbitrary and 21−α < 1 this is a contradiction. 2

More generally this prove works in any metric space that has “midpoints”, we willdiscuss these in more detail later. On bounded spaces (upper bound on all distances)and without midpoints, the notion of Holder α with α > 1 is less trivial.

One major point distinguishing uniform continuity and continuity is the followinglemma, which works only for uniformly continuous maps.

Lemma 33 Let (M,d) be a metric space and M ′ a subspace of M , and let (M, d)be some complete metric space. If f : M ′ → M is a uniformly continuous map,then there is a unique extension of this map from M to M which is also uniformlycontinuous.

Proof:Exercise: fix this proof:Let x ∈ M . Since g has a right limit of 0 at 0, for each dyadic r > 0 there exists

εr such that 2g(εr) + εr < r. Define

S = {(f(y), r) : (y, g(r)) ∈ B(x) ∩M ′}

Since M ′ is dense in M , this set is good.This set is a restrained set. namely, let (z, g(r)) in this set. Then z is of the form

f(y) with (y, g(r)) ∈ B(x), that is

lim infs→0

sup(u,s)∈S

d(y, u) + ε ≤ g(r)

We have to showlim infs→0

sup(v,s)∈S

d(z, v) + ε ≤ r

Since M is complete, the associated dedekind cone is of the form B(y) for somey ∈ M .

Define f(x) to be y.Then f thus defined is uniformly continuous (exercise) 2

A point x with f(x) = x is called a fixed point for a map x.

Lemma 34 (Contraction mapping principle) If f : M →M is a strict contrac-tion from a complete non-empty metric space to itself, then there exists a point x ∈Mwith f(x) = x

19

Proof: Since M is non-empty, there exists a point x0 in M . Assume we havedefined x1 already, then define xn+1 = f(xn).

We claim that for every n we have

d(xn+1, xn) ≤ Lnd(x1, x0)

For n = 0 this is clear since L0 = 1. Assume this inequality holds forNote that by the strict contraction property we have

d(xn+2), xn+1) = d(f(xn+1), f(xn)) =≤ Ld(xn+1), (xn)) ≤ Ln+1d(x1, x0)

Now we claim that there is a constant C such that for any n,m

d(xn, xn+m) ≤ C(2Ln − Ln+m)

This is clear for m = 0 as long as C is at least d(x0, x1). Assume we have proven thisfor m Then

d(xn, xn+m+1) ≤ d(xn, xn+m) + d(xn+m, xn+m+1) ≤ C(2Ln − Ln+m) + Ln+md(x0, x1)

= C(2Ln − Ln+m+1) + Ln+m(d(x0, x1)− C(L−1 − 1))

The last term is negative as long as C is larger than a number determined by L. Thiscompletes the induction.

Now since f is a contraction we have

d(f(x), x) ≤ d(f(x), f(xn)) + d(f(xn), xn) + d(xn, x) ≤ ε+ CLn + ε

Where ε and Ln can both be made arbitrarile small by choosing n large enough. Henced(f(x), x) = 0 and hence f(x) = x.

2

Exercise 17 The fix point guaranteed to exists in the above lemma is in fact unique.

The contraction mapping principle is a very ubiquitous method of choice to proveexistence for ODE or PDE. We discuss an example

Lemma 35 Let F be a Lipschitz map R → R and let c ∈ R. Then there exists asolution to the integral equation

f(x) = c+∫ x

0F (f(t)) dt

in C[0,∞).

20

Prove. We first construct ε > 0 such taht there is a solution to this equation inthe metric space C([0, ε]) with the supremum norm.

Define a map on this space by

T (f)(x) = c+∫ x

0F (f(t)) dt

This is a contraction mapping with respect to the supremumsFirst of all if f is continuosu, the F ◦ f is continuous since it is the composition

of continuous maps.The primitive of a continuosu function is continuous. Hence Tmaps M to itself.

We have, using that F is Lipshitz

d(T (f), T (g)) = supx

∣∣∣∣∫ x

0F (f(t)) dt−

∫ x

0F (g(t)) dt−

∣∣∣∣≤ d(T (f), T (g)) = sup

x

∫ x

0|F (f(t))− F (g(t))| dt

supx

∫ x

0|F (f(t))− F (g(t))| dt

supx

∫ x

0C |f(t)− g(t)| dt

supx

∫ x

0Cd(f, g) dt

εCd(f, g)

Hence this is a contraction mapping provided εC < 1 Hence there is a a continuousfunction f : [0, ε]→ R satisfying the equation.

We shall by induction prove that for every n that there exists continuous f on[0, nε, ] solving the equation. Assume this is shown for n . We shalls eek an extensionof f on [0, nε] to [0, (n+ 1)ε]. That satisfies for x ≥ n

f(x) = c+∫ x

0F (f(t)) dt

f(x) = c+∫ nε

0F (f(t)) dt+

∫ x

nεF (f(t) dt

This can be done by solving on [0, ε] the equation

g(x) = c+∫ nε

0F (f(t)) dt+

∫ x

0F (g(t) dt

and then setting f(x) = g(x− nε) Hiowever, to find a solution to the equation for gwe simply apply the previous argument.

2

Note that for c = 1 and F the identity map, this existence proof can be used toprove existence of the exponential function.

21

Lemma 36 Assume (K, d) is a compact metric space and (M,d′) is a metric space.Assume f : K →M is continuous. Then the range of f is compact.

Proof: Let C be an open cover of f(K). We produce a cover of K. Let x ∈ K.Let B′x be a ball in C that contains f(x). By continuity of f , there is a ball Bx inK such that f(Bx) ⊂ B′x. The collection of these balls Bx covers K. Pick a finitesubcover. Consider all B′x corresponding to the chosen Bx. This is a finite collection.We claim it covers f(K). Let y ∈ f(K). The there is z such that f(z) = y. Let Bx

be the chosen ball containing z. Then B′x contains y. 2

Lemma 37 Let (K, d) be a compact space and (M,d′) be a metric space. Let f :K →M be continuous. then f is uniformly continuous.

Fix s > 0. define for each x ∈ K r(x) to be the supremum of all numbers r ≤ 1such d(x, y) ≤ r implies d′(f(x), f(y)) ≤ δ. We claim that r is continuous, and firstsee how this will finish the proof. If continuous, then it attains its minimum on thecompact set K. The minimum cannot be 0 by continuity at that point. Hecne theminimu is larger than zero, which proves uniform continuity since s was arbitrary.

Now we turn to the proof of continuity. Fix x and note that in the vicinity oif xwe will use.

2

1.11 Contraction mapping principle

A point x with f(x) = x is called a fixed point for a map x.

Lemma 38 (Contraction mapping principle) If f : M →M is a strict contrac-tion from a complete non-empty metric space to itself, then there exists a point x ∈Mwith f(x) = x

Proof: Since M is non-empty, there exists a point x0 in M . Assume we havedefined x1 already, then define xn+1 = f(xn).

We claim that for every n we have

d(xn+1, xn) ≤ Lnd(x1, x0)

For n = 0 this is clear since L0 = 1. Assume this inequality holds for some n ≥ 0.Note that by the contraction property we have

d(xn+2, xn+1) = d(f(xn+1), f(xn)) ≤ Ld(xn+1), (xn)) ≤ Ln+1d(x1, x0)

Now we claim that there is a constant C such that for any n,m

d(xn, xn+m) ≤ C(2Ln − Ln+m)

22

This is clear for m = 0 as long as C is at least d(x0, x1). Assume we have proven thisfor some m ≥ 0. Then

d(xn, xn+m+1) ≤ d(xn, xn+m) + d(xn+m, xn+m+1) ≤ C(2Ln − Ln+m) + Ln+md(x0, x1)

= C(2Ln − Ln+m+1) + Ln+m(d(x0, x1)− C(L−1 − 1))

The last term is negative as long as C is larger than a number determined by L < 1.This completes the induction.

Now since f is a contraction we have

d(f(x), x) ≤ d(f(x), f(xn)) + d(f(xn), xn) + d(xn, x) ≤ εL+ 2CLn + ε

Where ε and Ln can both be made arbitrarily small by choosing n large enough.Hence d(f(x), x) = 0 and hence f(x) = x.

2

Lemma 39 The fix point guaranteed to exists in the above lemma is in unique.

Proof: assume that there are two fixed points x and y. Then d(f(x), f(y) = d(x, y)because both points are fixed points, but als d(f(x), f(y) ≤ Ld(x, y) because ofthe contraction property. Since L < 1, this is impossibly for d(x, y) 6= 0. Hecned(x, y) = 0 which implies x = y. 2

The contraction mapping principle is a very ubiquitous method of choice to proveexistence for ODE or PDE. We discuss an example

Lemma 40 Let F be a Lipschitz map R → R and let c ∈ R. Then there exists asolution to the integral equation

f(x) = c+∫ x

0F (f(t)) dt

in C[0,∞).

Prove. We first find small ε > 0 and a solution to this equation in the metricspace C([0, ε]) with the uniform metric given by supx∈[0,ε] |f(x)− g(x)|.

Define a map on this space by

T (f)(x) = c+∫ x

0F (f(t)) dt

We claim that this is a contraction mapping with respect to the uniform metric.First of all if f is continuous, then F ◦ f is continuous since it is the composition

of continuous maps. The primitive of a continuous function is continuous. Hence Tmaps M to itself.

We have, using that F is Lipschitz

d(T (f), T (g)) = supx

∣∣∣∣∫ x

0F (f(t)) dt−

∫ x

0F (g(t)) dt−

∣∣∣∣23

≤ d(T (f), T (g)) = supx

∫ x

0|F (f(t))− F (g(t))| dt

supx

∫ x

0|F (f(t))− F (g(t))| dt

supx

∫ x

0C |f(t)− g(t)| dt

supx

∫ x

0Cd(f, g) dt

εCd(f, g)

Hence this is a contraction mapping provided εC < 1. Hence choosing ε smaller than1/C there is a a continuous function f : [0, ε]→ R satisfying the equation.

We shall by induction prove that for every n ≥ 0 there exists a continuous f on[0, nε, ] solving the equation, the case n = 0 just having been established.

Assume this is shown for some n ≥ 0 . We shall seek an extension of f on [0, nε]to [0, (n+ 1)ε]. That extension needs to satisfy for x ≥ n

f(x) = c+∫ x

0F (f(t)) dt

f(x) = c+∫ nε

0F (f(t)) dt+

∫ x

nεF (f(t) dt

This can be done by solving on [0, ε] the equation

g(x) = c+∫ nε

0F (f(t)) dt+

∫ x

0F (g(t) dt

using the previous arguments and then setting f(x) = g(x− nε).2

Note that for c = 1 and F the identity map, this existence proof can be used toprove existence of the exponential function.

Exercise 18 (Newton method) Assume we have a function f : [0, 1] → R thatsatisfies for any x, y ∈ [0, 1] the Taylor expansion estimate

f(y)− f(x) + f ′(x)(y − x) ≤ Cx2

for some constant CDefine the iteration

xn+1 = xn −f(xn)

f ′(xn)

And prove that if x0 is within ε of a point x with f(x) = 0, then this iteration convergesto x.

This is called the Newton iteration method to find a zero.

24

1.12 Curves

We consider mappings f : [0, 1]→M with M a complete metric space.

Definition 18 An mapping f : [0, 1]→M is called

1. geodesic, if it is an isometry or equivalent to an isometry, that is of a becomesan isometry when the distance function on the target space is multiplied by anonzero constant.

2. a rectifiable curve, if it is Lipschitz

3. a curve if it is continuous (hence uniformly continuous)

Sometimes one restricts attention to injective mappins which are then called non-selfintersecting curves. We will however not need injectivity here.

We say two points x and y can be connected by a curve f if there is a curve withf(0) = x and f(1) = y.

Exercise 19 Prove that a subset of [0, 1] which contains the points 0 and 1, thesetwo points are connected by a curve if and only if it is the set [0, 1] itself.

Lemma 41 Let M be a metric space. Then the following two are equivalent:

1. For every x, y ∈ M there is a geodesic f : [0, 1] → M such that f(0) = x andf(1) = y.

2. For every x, y ∈ M there is a midpoint of x and y, that is a z ∈ M such thatd(x, z) = d(z, y) = d(x, y)/2

Proof: If x, y can be connected by a geodesic curve, then we pick such a geodesiccurve and define z = f(1/2). If λ is such that d(f(s), f(t)) = λd(s, t) for this geodesic,then

d(x, z) = λ/2 = d(x, z)/2

and similarly for d(z, y).Conversely, assume there is a midpoint for any two points in M . Assume without

loss of generality that d(x, y) = 1. Define a map f : [0, 1] → M as follows: Setf(0) = x and f(1) = y. Assume we have already defined f on Dn ∩ [0, 1]. Letz ∈ Dn+1 \Dn and let u ∈ Dn such that u ≤ z ≤ νn(u). Then define f(z) to be themidpoint of f(u) and f(νn(u)). By induction we see that d(t, t + νm(t)) = 2−m forx ∈ Dm. Let s < t be any two points in Dm, then we have by the triangle inequality

1 = d(f(0), f(1)) ≤ d(f(0), f(s)) + d(f(s), f(t)) + d(f(t), f(1))

≤∑

r∈Dm∩[0,1),r<sd(f(r), f(νm(r)))+

∑r∈Dm∩[0,1),s≤r<t

d(f(r), f(νm(r)))+∑

t∈Dm∩[0,1)d(f(t), f(νm(t)))

25

=∑ ∑

r∈Dm∩[0,1),r<s2−m

∑r∈Dm∩[0,1),s≤r<t

2−m∑

t∈Dm∩[0,1)2−m

= (s− 0) + (t− s) + (1− t) = 1

Since both sides of the inequality are 1, all intermediate inequalities in this argumenthave to be equalitites and in particular d(f(s), f(t)) = t− s. Hence f is an isometryand thus a geodesic. 2

Definition 19 We define the length of a curve to be

supn

∑x∈Dn,0≤x<1

d(f(x), f(νn(x))

Exercise 20 Let f : [0, 1]→M be a curve. Let g : [0, 1] ∩D→ [0, 1] be a monotonecurve with g(0) = 0 and g(1) = 1 (called a reparameterization). Then the length of fequals the length of f ◦ g.

Lemma 42 A curve has finite length if and only if there is a curve g onto the rangeof f which is Lipschitz.

Proof: Assume we have the map g such that h = f ◦g is rectifiable, with Lipschitzconstant L. Then we have for every n∑

x∈Dn,0≤x<1

d(h(x), h(νn(x))

≤∑

x∈Dn,0≤x<1

L2−n ≤ L

Conversely, assume the curve has finite length. Assume without loss of generalitythat this length is 1. Then we can define the arclength for every partial piece [0, x]to be

λ(x) = supn

∑t∈Dn,0≤t<x

d(f(t), f(νn(t))

Then λ is obviously a monotone function [0, 1] to [0, 1). We define a function g[0, 1]→[0, 1] by setting

g(λ(x)) = f(x)

To see that this is well defined, we need to verify that if λ(x) = λ(y) then f(x) = f(y).However, if λ(x) = λ(y), then by definiiton of λ in particular d(f(x), f(y)) = 0 becausethat term may appear in a sum that we take supremum over. Similarly we see thatg is Lipschitz, since we have for y > x

λ(x)− λ(y) ≥ d(f(x), f(y))

Clearly g is onto the range of f . 2

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Definition 20 A curve f : D ∩ [0, 1] → M is called of finite r-variation, if thereexists a constant C such that for any partition x0 < x1 < . . . < xn of [0, 1] (in Dm

say) we have

(n∑i=1

d(xi, xi+1)r)1/r ≤ C

Exercise 21 A curve f is of bounded r variation if ane only if there is a curveg : [0, 1] onto the range of f with which is Holder with exponent 1/r.

Definition 21 Let M be a complete metric space and x ∈ M . then the set of all ysuch that there exists a curve f with f(0) = x and f(1) = y is called the connectedcomponent Mx of x.

Definition 22 A metric space is called pathwise connected, if for any two points x, ythere exists a curve f with f(0) = x and f(1) = y. In other words, it is pathwiseconnected if Mx = M for every x ∈M .

Definition 23 A metric space is called locally pathwise connected, if for any point xthere exists a r > 0 so that Br(x) ⊂Mx.

Lemma 43 Assume M is locally pathwise connected. Then for every x the set Mx

is open.

Proof: Let y ∈Mx. Since M is locally pathwise connected there exists r such thatBr(y) ⊂My. It suffices to show that Br ⊂Mx.

Let z ∈ Br(y). Since also z ∈ My, there exist g : [0, 1] → M with g(0) = y andg(1) = z. Since y ∈Mx there exists f : [0, 1]→M with g(0) = y and g(1) = z.

Now define h : [0, 1]→M as follows

1. If t ≤ 1/2 then h(t) = f(2t)

2. If t ≥ 1/2 then h(t) = f(2t− 1)

This clearly defines a curve connecting x with z.

Lemma 44 Assume M is locally pathwise connected. Then for every x the set M\Mx

is open.

Proof: If y ∈ M \Mx, then also My ⊂ M \Mx. We prove this by contradiction,assume z ∈ My ∩Mx. Then we can connect z both with x and y, and then we canconstruct a curve from x to y as in the proof of the previous lemma. Since My isopen for every y ∈M \Mx, then M \Mx is open. 2

Definition 24 A metric space is called connected if it is not the union of two nonemptydisjoint open sets.

Lemma 45 A locally pathwise connected and connected set is pathwise connected.

Proof: For every x, since both Mx and M \Mx are open, one of them has to beempty. Since x ∈Mx, we have M \Mx = ∅.Exercise 22 Present and prove carefully an example for a connected metric sapcewhich is not pathwise connected.

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1.13 The space C(M,M ′)

Lemma 46 For two metric spaces (M,d) and (M ′, d′) consider the space (M, d) ofuniformly continuous functions f : M →M ′ with distance

d(f, g) = supx∈M

d(f(x), g(x))

Then this space is a complete metric space.

Note: this lemma works also with uniform continuity replaced by continuity.Proof: Suppose we have a Dedekind cone BinM . Then for every x ∈ M the set

Bx = {(f(x), r) : (f, r) ∈ B is a Dedekind cone in M .To see that, we need to show that for (y, r) ∈ Bx there exists ε with

infs

sup(y′,s)∈Bx

d′(y, y′) + ε ≤ r

Given such y, pick f ∈ B such that y = f(x). Choose ε as given by Dedekind coneproperty of B fro (f, r). Then

infs

sup(z,s)∈Bx

d′(f(x), z) + ε

= infs

sup(g,s)∈B

d′(f(x), g(x)) + ε

≤ infs

sup(g,s)∈B

d(f, g) + ε ≤ r

This proves that Bx is a Dedekind cone.Since M ′ is complete, there is a y ∈M ′ such that Bx = B(y).This gives a function g : M →M ′ assigning to each x this y.We claim that thsi function is uniformly continuous. For this we need to estimate

d′(g(x), g(x′)). This is the distance of two Dedekind cones, which by definition this is

infs

supf∈Bs

supf ′∈Bs

d(f(x), f ′(x′))

Fix s and poick an individual fs ∈ Bs. then we estimate

supf∈Bs

supf ′∈Bs

d(f(x), f ′(x′))

≤ supf∈Bs

supf ′∈Bs

(d(f(x), fs(x)) + d(fs(x), fs(x′)) + d(f ′(x′), f ′(x′)))

Now estimate sup by triangle inequality, giving three terms total. The first and lastterm are estimated alike, we discuss the first

supf∈Bs

supf ′∈Bs

d(f(x), fs(x))

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= supf∈Bs

d(f(x), fs(x))

= supf∈Bs

d(f, fs)

This can be estimated by 2s since B is a Dedekind cone.Now the middle term can be estimated by the modulus Gs of continuity of s

≤ supf∈Bs

supf ′∈Bs

d(fs(x), fs(x′))

≤ Gs(d(x, x′))

This proves uniform continuity of g.2

Note that the modulus of continuity of the function g produced was assembledby the modulus of continuity of a sequence of functions fs. Thus its modulus ofcontinuity may be worse than the modulus of continuity of either of these functions.This is why the corresponding sttement does nopt work for the class of Lipschitzfunctions, or the class of Holder functions. However, a similar theorem holds forcontinuous functions (we were only working locally at some x at a time, or say theclass of contraction mappings since then all functions along the way can be estimatedby the same modulus of continuity.

Exercise 23 The space of Lipschitz functions from M to M ′ is dense in the spaceC(M,M ′)

Exercise 24 The space of contractions from M to M ′ is complete in C(M,M ′)

1.14 Arzela Ascoli

Lemma 47 Let M and M ′ be compact spaces. Assume F ⊂ C(M,M ′). Assume thatthere is a modulus of continuity g such that all f ∈ F satisfy

g(d′(f(x), f(y)) ≤ d(x, y)

Then F is totally bounded.

Proof:Pick r > 0. Pick s = g(r/4). Cover M by finitely many balls Bs(xi), i = 1, . . . , n.

Cover the compact set M ′ by finitely many balls Br/4(yj), j = 1, . . .m.For each of the finitely many functions

h : {1, . . . , n} → {1, . . . ,m}

for which there exists a function f ∈ F with

f(xi) ∈ Br/4(yh(i))

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Pick one of these functions and call it fh. This is a finite collection in F . We claimthe corresponding balls of radius r about these functions in C(M,M ′) cover F .

To see this, let f be any function in F . Pick a function h such that

f(xi) ∈ Br/4(yh(i))

which exists by construction. We aim to show for all x ∈M that

d(f(x), fh(x)) ≤ r

Given such x, there is xi with x ∈ Bs(xi). Then

d(f(x), f ′(x)) ≤ d(f(x), f(xi)) + d(f(xi), yh(i)) + d(yh(i), fh(xi)) + d(fh(xi), fh(x))

But each of these numebers are less than r/4 by construction. This completes theproof.

2

1.15 The Riemann integral for continuous functions

So far we have defined the integral ∫ x

0f(t) dt

for monotone increasing functions. It satisfies∫ x

0(f(t) + g(t)) dt =

∫ x

0f(t) dt+

∫ x

0g(t) dt

and ∫ x

0f(t) dt ≤

∫ x

0g(t) dt

whenever f ≤ g.We can extend this to the space of fucntions of bounded variation. This is the set of

pairs (f, g) of monotone increasing functions with equivalence relation (f, g) ∼ (f ′, g′)if f + g′ = f ′ + g.

Recall that these functions take value in the real numbers, which is the set of pairs(x, y) with the equivalence relation (x, y) ∼ (x′, y′) if x+ y′ = x′ + y

The Riemann integral is then defined as∫ x

0(f(t), g(t)) dt = (

∫ x

0f(t) dt,

∫ x

0g(t) dt)

It does not depend on the choice of representative of the equivalence class, since forany two representatives (f, g), (f ′, g′) with f + g′ = f ′ + g we have∫ x

0f(t) dt+

∫ x

0g′(t) dt =

∫ x

0f(t) + g′(t) dt

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∫ x

0f ′(t) + g(t) dt =

∫ x

0f ′(t) dt+

∫ x

0g(t) dt

and hence(∫ x

0f(t) dt,

∫ x

0g(t) dt) ∼ (

∫ x

0f ′(t) dt,

∫ x

0g′(t) dt)

We refer to introduction of real numbers to elsewhere in the script and use theirproperties.

We like to extend this integral on BV ([0, x]) to the space C([0, x],R). Moreprecisely, neither of the spaces contains the other, even after natural identification of(f, g) ∈ BV ([0, x]) with the function f−g ∈ C([0, x]). However, the space of Lipschitzfunctions is in the inetrsection of both. Clearly Lipschitz functions are continuous.Moreover, if f is Lipschitz with constant L, then we can write it as difference of twpomonotone functions as follows:

f(t) = (f(t) + Lt)− Lt

The Lipschitz condition implies the first function is monotone. Recall that the Lips-chitz functions are dense in C[0, x]).

The key observation is that we have the continuity property (actual even a Lips-chitz continuity)

|∫ x

0f(t) dt−

∫ x

0g(t) dt| ≤ x‖f − g‖∞

This can be seen as follows. Let L be the maximum of the Lipschitz constants forf and g. Then we write

f(t) = (f(t) + L(t))− L(t)

g(t) = (g(t) + L(t))− L(t)

Representing each as monotone function. Then the LHS can be seen to be

|∫ x

0f(t) + Lt dt−

∫ x

0g(t) + Lt dt|

Denote ‖f − g‖∞ by d, then∫ x

0f(t) + Lt dt ≤

∫ x

0g(t) + Lt+ d dt =

∫ x

0g(t) + Lt dt+ xd

∫ x

0g(t) + Lt dt ≤

∫ x

0f(t) + Lt+ d dt =

∫ x

0f(t) + Lt dt+ xd

This proves the desired Lipschitz continuity.Note that this continuity property holds on all of BV .Since Lipschitz functions are dense in C([0, x]), there is a unique extension to

C([0, x]) if this Lipschitz mapping. This defines∫ x

0f(t) dt

for functions f in C([0, 1])

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Similarly one can see that the n-th left and rigth Riemann sums can be extendedto C([0, 1]). Moreover, all the n-th Riemann sums for a fixed function are boundedby ‖f‖∞x and Lipschitz with constant x. This makes them an equicontinuous family.

On the compact space of contarctions, the left and right Riemann sums have aconvergent subsequecne, which converges to the integral since Liipschitz functiosn areBV. Sicne the limit is independent of the supsequence, the whole sequence converegsto this limit.

By scaling the same holds for Lipschitz functions. Since Lipschitz functions aredense, the same holds for all C([0, 1],R) functions.

Let I : C([0, 1])→ C([0, 1]) be defined by

If(x) =∫ x

0f(t) dt

First of all, we note that indeed If defines a Lipschitz function. Namely, if x < y wehave

If(y)− If(x) =∫ y

xf(t) dt ≤ |y − x|‖f‖∞

Hence If is even a Lipschitz function.We like to understand properties of the range of the map I itself.

Definition 25 We call a Lipschitz function g : [0, 1]→ R continuously differentiable,if there is a monotone decreasing sequence εn with limit 0 such that for each x ∈ [0, 1)the functions

g(x+ εn)− g(x)

εn

converge uniformly to a function g which is a uniformly continuous function on [0, 1).(Note that for given x, the terms in this sequence are only defined for large enoughn.)

Lemma 48 The range of I is contained in the space of continuously differentiablefunctions which vanish at 0. I.e., If f is continuous and g = If , then for eachx ∈ [0, 1) we have

limn→∞

If(x+ 2−n)− If(x)

2−n= f(x)

Proof: Note if n is large enough then

εn(f(x)− ε) ≤ If(x+ εn)− If(x) ≤ εn(f(x) + ε)

This proves existence of the limit. Since f is uniformly contiunuous, this proves thelemma. 2

Lemma 49 The map I is injective.

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Proof: Suppose If = Ig. Then I(f − g) = 0. Hence it suffices to show Ih = 0implies h = 0. Assume to get a contradictiopn h 6= 0, then there exists y ∈ (0, 1)with h(y) 6= 0. Assume wlog h(y) > ε > 0. Then there is a ball of radius δ about ywith h(x) > ε in this ball. But then

Ifh(y − δ) + 2εδ ≤ Ih(y + δ)

And thus Ih 6= 0. This is a contradiction. 2

Lemma 50 The map I is surjective onto the space of continuously differentiablefunctions vanishing at the origin.

Proof: Let g be some continuously differentiable function, and let f denote itscontinuous derivative (extended to [0, 1]). It suffices to show g = If .

Consider ∫ x

0

g(t+ εn)− g(t)

εndt

=1

εn

∫ x

0g(t+ εn) dt− 1

εn

∫ x

0g(t) dt

=1

εn

∫ x+εn

εng(t) dt− 1

εn

∫ x

0g(t) dt

=1

εn

∫ x+εn

xg(t) dt− 1

εn

∫ ε

0g(t) dt

Now the first term tends to g(x), while the secodn term tends to 0 since g vanishesat 0. Hence

limn→∞

∫ x

0

g(t+ εn)− g(t)

εndt = g(x)

It remains to interchange the limit with the integral on the left hand side. However,this follwos from uniform convergence of the integrand. (It follows also by Lebesguedomeinated convergence, without having to demand uniform converegcen of the in-tegrand, but we do not want to use that here.

2

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